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Exciting a photon

by fouad89
Tags: exciting, photon
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Simon Bridge
Jan25-13, 08:07 PM
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Quote Quote by soarce View Post
I wouldn't take too seriously the models, it doesn't have to be associated to a real process. One use model to calculate things and compare them with the experiments.
Most people would hope that the model does have to have some relationship to a real process otherwise, how can you claim to understand them? But this is not the place to debate philosophy of science.
Anybody measured the short interval between absorption and re-emission of the photon?
Yes. The mean times are of order of ##10^{-23}s## for the photon scattering off an electron (I'd be hard-pressed to locate the paper though) - but it can be quite long depending on the energy of the photon and the situation the electron is in.
All those processes involve virtual states and, as far I know, the scattering take place instantaneously.
In the model, it just takes a very short time compared to the rest of the diagram - so the lines are horizontal - but see the vertical lines too?. Of course there is this issue about whether the virtual particles have a physical existence when they are mediating an interraction like this or if they are just an artifact of the math... and it's more like the wave-function has a spread in space rather than that the particle translates classically. For the photon scattering off an electron, it is a real, physical, electron all the way through.
The picture that the photon is absorbed, the system holds on for a while and then re-emits the photon is wrong.
Recall that the interactions at the scale of photons are supposed to be local - no "action at a distance". So the other way for an electron to scatter a photon is via a virtual photon of it's own - and the Feynman diagrams for that sum to zero. So what were you thinking happens?

By "exciting a photon" he probably ment giving energy of a photon.
That's certainly one interpretation.
If you have given energy to a photon, is it analogous to giving energy to an electron?
Is it analogous to "exciting" and electron?

Seems an odd way to phrase it (post #1) is that was what was meant - but it certainly could be the case. I suppose it is up to OP to clarify what was intended.

Either way, the question has been answered ;)

I agree. Based on Feynman diagram model the photon itself may go through a virtual electron-positron pair... It maintains its identity or it keeps changing ? :)
That's the philosophical part... if you get from one place to another by being destroyed and recreated - to what extent is it reasonable to say it is still you? If the exiting photon is identical (same energy and momentum, and spin) then there is probably a case for, at least, treating it as the same photon. i.e. in classical reflection, we treat the light reflected off the mirror as being the same light that was incident to it a moment earlier. At the photon level, though, the law of reflection is only obeyed on average even, so these kinds of things get tricky.

Quote Quote by Naty1
...but like it or not, a diminishing wave function outside a finite potential well does interest is whether it has any predictable physical effects, and of course whether they can be experimentally verified at sometime. In other words, which math, which models, fit our universe....
I suspect that is outside the scope of this thread. So just quickly: the extent of the wavefunction outside the classical limits does predict real world effects - like tunneling. It also changes the predicted energy-levels and thus the material properties.

You'll notice that I answered a slightly different question to the one asked though.


To summarize:
If we read the question in terms of changing the energy of a photon via some interractions, then there are several ways this may be done. It would be analogous to accelerating an electron.

If we read it in terms of bound-particle quantum states, analogous to "exciting" and electron (or an atom) then this is not so clear cut ... the simple answer would be "no".

There is some issue around whether you can legitimately call the final photon "the same photon" as the initial one ... depending on the details of the situation. The resolution would be up to the model you want to use.

I think, between all the posts, we've covered the possible misunderstandings :)
Remains only to get feedback from the OP :D <waves>

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